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Moment of inertia of a rod with balls attached

  1. Aug 7, 2017 #1
    1. The problem statement, all variables and given/known data
    Attached.

    2. Relevant equations
    I=mr^2, I=(1/12)m*l^2 for a rod.

    3. The attempt at a solution
    Part A I got by doing I=(1/12)3.5*(2.6)^2+2*.7*1.3^2 (I added the moment of inertia of the rod and the balls). Part B since the axis is on one of the balls, I thought we don't include that ball in the calculations. So the rod stays, (1/12)m*l^2 and we add the other .7 kg ball which is 2.6m away. So I_total=(1/12)3.5*2.6^2+.7(2.6)^2=6.7 kg*m^2. However that isn't the right answer. Can someone please explain to me where I went wrong. Any help is greatly appreciated. Physics Q.png
     
  2. jcsd
  3. Aug 7, 2017 #2

    TSny

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    Think about that.
     
  4. Aug 7, 2017 #3
    Can you explain your hint. Why wouldn't you include the I of the rod?
     
  5. Aug 7, 2017 #4

    TSny

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    You do need to include I for the rod. But when you move the axis of rotation to one end of the rod, does I for the rod remain the same?
     
  6. Aug 7, 2017 #5
    Im confused. The only reason why I knew I=1/12ml^2 is because I looked up moment of inertia for a rod. The mass is the same and the length of the rod is the same? Can you please explain what it would change to and why?
     
  7. Aug 7, 2017 #6

    TSny

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    Moment of inertia depends on how the mass of the object is distributed around the axis of rotation. When you switch the axis, you change how the mass is distributed relative to the new axis. Consult PAT (Parallel Axis Theorem).

    http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html
     
  8. Aug 7, 2017 #7
    Alright so I'm using the theorem. So I_total(rod)=I_cm+md^2=(1/12)3.5*2.6^2+4.9(1.3)^2=10.25
    10.25+.7(2.6)^2=14.98 kg*m^2. Is this right?
     
  9. Aug 7, 2017 #8

    TSny

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    It looks OK except for the factor of 4.9. How did you get that number?
     
  10. Aug 7, 2017 #9
    It's the M in the parallel axis equation which I thought meant the sum of all the masses.
     
  11. Aug 7, 2017 #10

    TSny

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    You should apply the theorem to just the rod. You are taking care of the point masses separately.
     
  12. Aug 7, 2017 #11
    I_total(rod)=I_cm+md^2=(1/12)3.5*2.6^2+3.5(1.3)^2=7.89
    7.89+.7(2.6)^2=12.62 kg*m^2. I just put it in and its right, thanks a lot for your help Tsny.
     
  13. Aug 7, 2017 #12

    TSny

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    Great! Good work.
     
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